Spectral density and Wiener–Khinchin
Move between covariance and frequency using an explicit radians convention, while keeping spectral atoms distinct from a density.
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The question for today
What you’ll understand
- Use the angular-frequency Fourier pair with the correct 2π factors.
- Distinguish a general spectral measure from an absolutely continuous density.
- Calculate ARMA and filtered spectra as variance allocations across frequency.
Covariance describes how observations co-vary across lags; a spectrum describes how their variance is distributed across frequencies. These are two views of the same second-order structure, provided the Fourier convention is fixed. A stationary process always has a finite spectral measure, but that measure may contain atoms and need not have a density. Even when a density exists, switching between cycles and radians changes its numerical scale. Ignoring that factor can make an otherwise correct spectral calculation integrate to the wrong variance.
Fix radians per observation
For a centered real weakly stationary process, take gamma(h)=Cov(X at t+h,X at t). If the covariance sequence is absolutely summable, its Fourier series defines a continuous nonnegative even density on minus pi to pi. We put one over two pi in the forward transform and no prefactor in the inverse integral. Therefore the integral of the density equals gamma(0), the variance. Frequencies are radians per observation, not cycles per unit of physical time.
Many time-series references instead use cycles nu between minus one half and one half, with the exponent minus two pi i nu h. Their cycles density S(nu) equals two pi times our angular density evaluated at two pi nu. In particular, weak white noise of variance sigma-squared has cycles density sigma-squared and angular density sigma-squared divided by two pi. These are the same variance allocation after the change of variable in the integral.
A measure can carry variance on spectral lines
Without the summability assumption, the positive-semidefinite covariance sequence still has a finite nonnegative spectral measure by Herglotz's theorem. The inverse uses integration with respect to that measure. A random constant of variance v has covariance v at every lag and a spectral atom of mass v at zero; it has no ordinary density representing that mass. Random sinusoidal components similarly give spectral lines at paired nonzero frequencies under suitable stationary amplitude laws.
The measure is not a probability distribution unless normalized by the total variance. Its density, when present, can vanish at some frequencies without invalidating the process. Nor does having an absolutely continuous spectral measure by itself establish pure nondeterminism: the innovation factorization has additional regularity requirements. Wold's linear-past condition and a graph that looks smooth in frequency should therefore not be treated as interchangeable definitions.
Low-frequency strength is a covariance pattern
A stable positive AR(1) coefficient produces a density concentrated toward zero frequency, reflecting slowly decaying positive covariances. A negative coefficient shifts emphasis toward pi, reflecting alternating lag correlations. These are properties of a stationary process and do not by themselves imply a deterministic trend. A near-unit root can create a very sharp stationary low-frequency peak, whereas a genuine random walk has no stationary covariance density of the form used here.
Differencing multiplies a stationary input spectrum by the squared modulus of one minus exp of minus i omega, suppressing zero frequency. It can be an appropriate transformation for integrated data, but applying it to already stationary white noise creates a negatively correlated MA process. Spectral filtering changes the information available for a task; it is not automatically a cure for every visual trend or a proof of a causal relation between two series with similar peaks.
Work it through
Check an angular AR spectrum against its variance
For X_t=0.5X_{t−1}+epsilon_t with unit-variance weak white noise, compute f(0), f(pi), their ratio and total integrated variance.
The denominator is |1−0.5e^(−i omega)|²=1.25−cos omega, so f=1/[2pi(1.25−cos omega)].
At zero the denominator inside the parentheses is one quarter, giving f(0)=2/pi. At pi it is nine quarters, giving f(pi)=2/(9pi), a ratio of nine.
The inverse transform at lag zero integrates to the known stationary variance 1/(1−0.5²)=4/3. Equivalently, expanding the causal geometric filter gives this sum of squared coefficients.
The angular density integrates to 4/3. Omitting its one-over-two-pi factor would instead inflate the integrated variance by two pi.
The living notebook
Explore the model
Interactive lab · waves
Square the transfer amplitude to compare power
Pause and reset, then set both controls to phase difference zero degrees and amplitude ratio one. Predict summed amplitude two and mean squared amplitude four times that of either component. For a fresh comparison, pause and reset, then set phase difference 180 degrees and ratio one; predict cancellation to numerical roundoff. These are coherent fixed-frequency sinusoids, illustrating the squared-modulus transfer factor. The instrument is not a stochastic spectral-density estimator and does not make independent noise components coherent.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forFiltering acts on amplitudes through a complex transfer function and on spectral power through its squared modulus.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let X be unit-variance weak white noise and set Y_t=X_t−X_{t−1}. Find its angular spectrum and its nonzero autocovariances. Is differencing this stationary input guaranteed to leave white noise?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use |1−e^(−i omega)|²=2−2cos omega, or expand the covariance directly.
Compare with the solution
The spectrum is (1−cos omega)/pi. Its integral is two, agreeing with gamma-Y(0)=2. Adjacent differences share one original noise term with opposite signs, giving gamma-Y(±1)=−1, and the remaining nonzero-lag covariances vanish. Thus the output is stationary but not white. It is an MA(1) with coefficient minus one, on the boundary of invertibility, illustrating why differencing an already stationary series can introduce artificial negative short-lag dependence.
Keep the idea
Take it with you
- Declare radians or cycles and check that the spectrum integrates to variance.
- Stationarity guarantees a spectral measure; a density requires additional absolute continuity.
- A linear filter multiplies spectral power by the squared modulus of its transfer function.
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