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Stationary Processes and Dependence · Tutorial 362 of 952

Spectral Density as the Fourier Transform of Autocovariance

Learn how autocovariances determine a process’s spectral density, why the density is symmetric and nonnegative, and how to calculate it for an AR(1).

Beginner 9 min read

What You'll Learn

  • Define the spectral density using the Fourier transform of an absolutely summable autocovariance sequence.
  • Recover autocovariances from the spectral density using Fourier inversion.
  • Explain why a spectral density is nonnegative and even for a real-valued covariance-stationary process.
  • Calculate and interpret a spectral density for a finite-lag autocovariance sequence.
  • Derive and evaluate the spectral density of a stationary AR(1) process.

From Autocovariance to Frequency

A covariance-stationary process can be described through how its observations covary across time. The autocovariance at lag \(h\) records dependence between observations \(h\) periods apart. The spectral density expresses the same second-order information in the frequency domain: instead of organizing variation by time separation, it organizes it by frequency.

This connection is a Fourier transform. Its usefulness depends on being precise about the convention, including the factor \(1/(2\pi)\). Throughout, let \(\{X_t\}\) be a real-valued covariance-stationary process with mean \(\mu\), and define its autocovariance by \(\gamma(h)=\operatorname{Cov}(X_t,X_{t-h})\). Assume that its autocovariances are absolutely summable:

$$ \sum_{h=-\infty}^{\infty}|\gamma(h)|<\infty. $$

This condition ensures that the Fourier series below converges absolutely and uniformly. Frequencies are measured in radians per observation and are considered on \([-\pi,\pi]\), with endpoints equivalent by periodicity.

Definition: Under absolute summability, the spectral density of \(\{X_t\}\) is
$$ f(\omega)=\frac{1}{2\pi}\sum_{h=-\infty}^{\infty}\gamma(h)e^{-ih\omega}, \qquad -\pi\leq\omega\leq\pi. $$
The corresponding Fourier inversion formula is
$$ \gamma(h)=\int_{-\pi}^{\pi}e^{ih\omega}f(\omega)\,d\omega. $$

The normalization is part of the definition used here. In particular, integrating the density over the full frequency interval gives the variance:

$$ \int_{-\pi}^{\pi} f(\omega)\,d\omega=\gamma(0)=\operatorname{Var}(X_t). $$

Thus the spectral density is scaled so that its total area, not its average height, equals the process variance. Since radians are dimensionless mathematically, its units can be described as squared units of \(X_t\) per unit frequency.

Fourier Inversion and the Real-Valued Case

The inversion formula follows from the orthogonality of complex exponentials. Absolute summability permits integrating the defining series term by term. Multiplying by \(e^{ih\omega}\) gives

$$ \begin{aligned} \int_{-\pi}^{\pi}e^{ih\omega}f(\omega)\,d\omega &=\frac{1}{2\pi}\sum_{j=-\infty}^{\infty}\gamma(j) \int_{-\pi}^{\pi}e^{i(h-j)\omega}\,d\omega\\ &=\frac{1}{2\pi}\sum_{j=-\infty}^{\infty}\gamma(j)\,2\pi\mathbf{1}\{j=h\}\\ &=\gamma(h). \end{aligned} $$

Here the integral of \(e^{ik\omega}\) over \([-\pi,\pi]\) is \(2\pi\) if the integer \(k=0\), and zero otherwise. Setting \(h=0\) yields the variance identity.

For a real-valued covariance-stationary process, \(\gamma(-h)=\gamma(h)\). Consequently, paired positive and negative lags turn the complex Fourier series into a real cosine series:

$$ f(\omega)=\frac{1}{2\pi} \left[\gamma(0)+2\sum_{h=1}^{\infty}\gamma(h)\cos(h\omega)\right]. $$

It follows directly that \(f(-\omega)=f(\omega)\): the spectral density is even. The exponential form is often convenient for derivations, while the cosine form makes the real-valued symmetry explicit.

Worked Example: A Finite-Lag Autocovariance Sequence

Suppose a hypothetical process has autocovariances \(\gamma(0)=5\), \(\gamma(1)=\gamma(-1)=1\), and \(\gamma(h)=0\) for \(|h|>1\), in squared measurement units. Find its spectral density and compare its values at zero and at \(\pi\).

Only three terms in the Fourier series are nonzero:

$$ \begin{aligned} f(\omega) &=\frac{1}{2\pi}\left(5+e^{-i\omega}+e^{i\omega}\right)\\ &=\frac{5+2\cos(\omega)}{2\pi}. \end{aligned} $$

At zero frequency, \(\cos(0)=1\), so

$$ f(0)=\frac{7}{2\pi}\approx 1.114. $$

At frequency \(\pi\), \(\cos(\pi)=-1\), giving

$$ f(\pi)=\frac{3}{2\pi}\approx 0.4775. $$

The density is nonnegative at every frequency because its numerator ranges from 3 to 7. Its total area is \(\gamma(0)=5\), as required by Fourier inversion. These checks are useful when translating a finite list of autocovariances into a density.

Why the Spectral Density Cannot Be Negative

The formula alone does not make nonnegativity obvious: individual autocovariances can be positive or negative, and cosine terms can have either sign. The key fact is that autocovariances come from variances of linear combinations of observations, and variances cannot be negative. The argument below connects that fact to the Fourier transform.

Let \(Y_t=X_t-\mu\), and for a fixed frequency \(\omega\), form the complex-valued finite sum \(S_N(\omega)=\sum_{t=1}^{N}Y_t e^{-it\omega}\). Its squared magnitude has nonnegative expectation. Grouping terms by the lag between observations gives

$$ \begin{aligned} \frac{1}{N}E|S_N(\omega)|^2 &=\sum_{|h|<N}\left(1-\frac{|h|}{N}\right)\gamma(h)e^{-ih\omega}\\ &\geq 0. \end{aligned} $$

For each lag \(h\), there are \(N-|h|\) pairs of observations in the sum whose separation is \(h\), which explains the weight \(1-|h|/N\). As \(N\) increases, this weight tends to 1 for each fixed lag. Absolute summability allows the limit to pass through the series: the terms are bounded in absolute value by \(|\gamma(h)|\), whose sum is finite. Therefore,

$$ \lim_{N\to\infty}\frac{1}{N}E|S_N(\omega)|^2 =\sum_{h=-\infty}^{\infty}\gamma(h)e^{-ih\omega} =2\pi f(\omega)\geq 0. $$

This proves nonnegativity at every frequency. The complex sum in the argument is only a convenient mathematical device; for a real-valued process, the resulting density is real and even.

Theorem: If a real-valued covariance-stationary process has an absolutely summable autocovariance sequence, then its spectral density, defined by the Fourier transform above, is real, even, and nonnegative. Its integral over \([-\pi,\pi]\) equals the process variance.

Working with a Summable Geometric Sequence

Absolute summability also makes familiar infinite-series calculations available. The following hypothetical autocovariance sequence illustrates how a Fourier series can be simplified before evaluating it at particular frequencies.

Worked Example: Summing a Geometric Autocovariance Sequence

Consider the sequence \(\gamma(h)=2(1/3)^{|h|}\) for integer \(h\). Its absolute sum is finite:

$$ \sum_{h=-\infty}^{\infty}|\gamma(h)| =2\left(1+2\sum_{h=1}^{\infty}(1/3)^h\right) =2(1+1)=4. $$

Write \(q=1/3\). Pairing positive and negative lags and summing the geometric series gives

$$ \begin{aligned} \sum_{h=-\infty}^{\infty}\gamma(h)e^{-ih\omega} &=2\left[1+2\sum_{h=1}^{\infty}q^h\cos(h\omega)\right]\\ &=2\frac{1-q^2}{1-2q\cos(\omega)+q^2}. \end{aligned} $$

At \(\omega=\pi/2\), the cosine is zero. Substitution gives

$$ \begin{aligned} \sum_{h=-\infty}^{\infty}\gamma(h)e^{-ih\pi/2} &=2\frac{1-1/9}{1+1/9} =1.6,\\ f(\pi/2)&=\frac{1.6}{2\pi}\approx 0.2546. \end{aligned} $$

At zero frequency, the same expression is 4, so \(f(0)=4/(2\pi)=2/\pi\approx0.6366\). At \(\pi\), it is 1, so \(f(\pi)=1/(2\pi)\approx0.1592\). The density is even, and its integral is \(\gamma(0)=2\).

The Spectral Density of a Stationary AR(1)

Consider the stationary AR(1) process

$$ X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t, \qquad |\phi|<1, $$

where the innovations \(\varepsilon_t\) have mean zero, variance \(\sigma_\varepsilon^2\), and are uncorrelated across time. Under \(|\phi|<1\), the stationary linear-process representation gives the autocovariance sequence

$$ \gamma(h)=\frac{\sigma_\varepsilon^2}{1-\phi^2}\phi^{|h|}. $$

Because \(|\phi|<1\), this sequence is absolutely summable. Substituting it into the Fourier formula and summing the two geometric series yields

$$ \begin{aligned} f(\omega) &=\frac{1}{2\pi}\frac{\sigma_\varepsilon^2}{1-\phi^2} \left[1+\sum_{h=1}^{\infty}\phi^h e^{-ih\omega} +\sum_{h=1}^{\infty}\phi^h e^{ih\omega}\right]\\ &=\frac{\sigma_\varepsilon^2}{2\pi} \frac{1}{1+\phi^2-2\phi\cos(\omega)}. \end{aligned} $$

The denominator is \(|1-\phi e^{-i\omega}|^2\), and is strictly positive for \(|\phi|<1\). Thus the formula is consistent with the general nonnegativity result.

Worked Example: Evaluating an AR(1) Spectrum

Suppose a hypothetical stationary AR(1) has \(\phi=0.6\) and innovation variance \(\sigma_\varepsilon^2=4\) squared measurement units. Its spectral density is

$$ f(\omega)=\frac{4}{2\pi(1+0.6^2-1.2\cos(\omega))} =\frac{4}{2\pi(1.36-1.2\cos(\omega))}. $$

At zero frequency, the denominator inside the parentheses is \(1.36-1.2=0.16\), so

$$ f(0)=\frac{4}{2\pi(0.16)} =\frac{12.5}{\pi} \approx 3.979. $$

At \(\pi/2\), \(\cos(\pi/2)=0\), giving

$$ f(\pi/2)=\frac{4}{2.72\pi}\approx 0.4681. $$

At \(\pi\), the denominator inside the parentheses is \(1.36+1.2=2.56\), hence

$$ f(\pi)=\frac{4}{5.12\pi}\approx 0.2487. $$

These values are squared measurement units per unit frequency. They illustrate the shape of this particular density: it is largest at zero frequency and smaller at \(\pi\). The density is even, so the value at \(-\pi/2\) is also approximately \(0.4681\).

Interpretation and a Common Pitfall

A spectral density is another representation of second-order dependence, not a separate probability distribution. Its values describe how the covariance structure is allocated across frequencies, while its area recovers the variance. It is therefore important to keep the definition’s scale and frequency interval in view when calculating or comparing spectral densities.

  • Omitting the normalization. Using \(\sum_h\gamma(h)e^{-ih\omega}\) as the density without dividing by \(2\pi\) changes the convention and makes the integral over \([-\pi,\pi]\) equal \(2\pi\gamma(0)\), not \(\gamma(0)\).
  • Assuming negative covariance terms imply a negative density. Individual terms in the Fourier series may be negative, but the full sum is nonnegative because it is the limit of scaled expected squared magnitudes.
  • Forgetting symmetry. For a real-valued covariance-stationary process, positive and negative lags pair to produce cosines, so \(f(-\omega)=f(\omega)\).
  • Reading density height as variance. A value \(f(\omega)\) is not the variance by itself. The variance is the area under the density over the full interval.

A reliable calculation begins by stating the Fourier convention, checking that the autocovariances are absolutely summable, and then checking symmetry, nonnegativity, and the variance integral. These checks catch many algebra and normalization errors.

Check Your Understanding

Use the convention and definitions in this tutorial to answer the following.

  1. Why does absolute summability of the autocovariances make the Fourier-series definition well behaved?
  2. For a real-valued covariance-stationary process, explain why its spectral density is even.
  3. What variance quantity is recovered by integrating the spectral density over \([-\pi,\pi]\)?
  4. For the finite-lag sequence in the first worked example, why is its spectral density nonnegative at every frequency?
  5. For the AR(1) density, identify the condition on \(\phi\) that ensures the geometric autocovariance sequence is absolutely summable.
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