Spectral Densities and Fourier Inversion

From Spectral Measure to Spectral Density

A spectral measure distributes the variance of a weakly stationary process across frequencies. When that measure has a density, integration against the density recovers each covariance. The normalization matters: with angular frequency on $[-\pi,\pi)$ and ordinary Lebesgue measure $d\lambda$, white noise of variance $\sigma^2$ has constant density $\sigma^2/(2\pi)$, not $\sigma^2$.

Let $(X_n)_{n\in\mathbb Z}$ be a real-valued process on a probability space $(\Omega,\mathcal F,\mathbb P)$, with state space $S=\mathbb R$ and $\mathbb E[X_n^2]<\infty$. Assume it is weakly stationary: its mean is constant, and its covariance depends only on the lag. Write $\gamma(h)=\operatorname{Cov}(X_{n+h},X_n)$, for $h\in\mathbb Z$. Its spectral measure $F$ is a finite positive measure on the frequency circle, represented by $[-\pi,\pi)$ with the endpoints identified, under the convention

\gamma(h)=\int_{[-\pi,\pi)} e^{ih\lambda}\,F(d\lambda), \qquad h\in\mathbb Z.

This is the angular-frequency convention for the spectral measure. In particular, $F([-\pi,\pi))=\gamma(0)=\operatorname{Var}(X_n)$. The existence of such a finite positive measure for a weakly stationary covariance sequence is the standard Herglotz theorem from the earlier material on spectral measures.

Definition: The spectral measure $F$ is absolutely continuous with respect to Lebesgue measure if every Lebesgue-null set has $F$-measure zero. In that case its spectral density is a nonnegative integrable function $f$ satisfying $F(B)=\int_B f(\lambda)\,d\lambda$ for every measurable set $B\subseteq[-\pi,\pi)$. The density is determined only up to changes on a Lebesgue-null set.

The phrase “has a spectral density” means that the entire spectral measure is represented by this integral. If a spectral measure has a part that is not absolutely continuous, a density for its absolutely continuous part alone does not recover the entire covariance sequence. No continuity or boundedness of $f$ is required: integrability and nonnegativity almost everywhere are enough.

Fourier Inversion for Covariances

Theorem (Covariance inversion from a density): Suppose a weakly stationary process has spectral measure $F$ under the convention above, and $F$ is absolutely continuous with spectral density $f$. Then $f\geq 0$ almost everywhere, $\int_{-\pi}^{\pi} f(\lambda)\,d\lambda=\gamma(0)$, and for every integer $h$, $$ \gamma(h)=\int_{-\pi}^{\pi} e^{ih\lambda}f(\lambda)\,d\lambda. $$ For a real-valued process, $\gamma(-h)=\gamma(h)$; correspondingly, its spectral measure is symmetric, and its density can be chosen even almost everywhere.

Proof. Absolute continuity means, by the definition of the density, that $F(d\lambda)=f(\lambda)\,d\lambda$. Since $F$ is a positive measure, its density is nonnegative almost everywhere. Taking $h=0$ in the spectral-measure representation gives $\gamma(0)=F([-\pi,\pi))$. The density definition then gives $\gamma(0)=\int_{-\pi}^{\pi}f(\lambda)\,d\lambda$; changing the value at the single identified endpoint does not affect this Lebesgue integral. For any integer $h$, substitute $F(d\lambda)=f(\lambda)\,d\lambda$ into the representation of $\gamma(h)$ to obtain $\gamma(h)=\int_{-\pi}^{\pi}e^{ih\lambda}f(\lambda)\,d\lambda$. Finally, for a real process, covariance symmetry gives $\gamma(-h)=\gamma(h)$. The uniqueness of the spectral measure in the Herglotz representation implies that reflecting its frequencies leaves the measure unchanged; its density is therefore equal to its reflection almost everywhere. $\square$

The integral is a Fourier coefficient of $f$, with the sign fixed by the stated convention. It directly recovers each covariance; it does not assert that a density can be recovered pointwise from an arbitrary covariance sequence by summing a Fourier series. Pointwise Fourier-series inversion needs additional hypotheses and is a different question.

A useful converse check is that a proposed density cannot generate negative variances or an invalid covariance sequence. For an even density, this follows from positivity of the measure it defines.

Proposition: Let $f$ be a nonnegative, integrable, even function on $[-\pi,\pi)$, with the endpoints identified. Define $\gamma(h)=\int_{-\pi}^{\pi}e^{ih\lambda}f(\lambda)\,d\lambda$. Then $\gamma(0)=\int f$, $\gamma(h)$ is real and satisfies $\gamma(-h)=\gamma(h)$, and $\gamma$ is positive definite: for every integer $m\geq 1$, every $n_1,\ldots,n_m\in\mathbb Z$, and every $c_1,\ldots,c_m\in\mathbb C$, $$ \sum_{r=1}^{m}\sum_{s=1}^{m}c_r\overline{c_s}\,\gamma(n_r-n_s)\geq 0. $$

Proof. At lag zero, the exponential is $1$, so $\gamma(0)=\int f$. Since $f$ is real and even, the sine part of $\int e^{ih\lambda}f(\lambda)\,d\lambda$ integrates to zero, while the cosine part is real and unchanged when $h$ is replaced by $-h$. For positive definiteness, use the defining integral and finite-sum linearity to write $\sum_{r,s}c_r\overline{c_s}\gamma(n_r-n_s)=\int f(\lambda)\sum_{r,s}c_r\overline{c_s}e^{i(n_r-n_s)\lambda}\,d\lambda$. The inner sum equals $\left|\sum_r c_r e^{in_r\lambda}\right|^2$. Thus the expression is $\int f(\lambda)\left|\sum_r c_r e^{in_r\lambda}\right|^2\,d\lambda$, which is nonnegative because $f\geq0$ almost everywhere. This proves all the claims. By the standard Herglotz theorem, this positive-definite sequence is a valid covariance sequence. $\square$

Worked Covariance Integrals

Worked Example: Constant Density and White Noise

Suppose a weakly stationary real process has spectral density $f(\lambda)=\tfrac{3}{2\pi}$ on $[-\pi,\pi)$. Recover its covariance at every integer lag.

First check the total mass: $\int_{-\pi}^{\pi}\tfrac{3}{2\pi}\,d\lambda=\tfrac{3}{2\pi}(2\pi)=3$, so the variance is $3$. At lag zero, $\gamma(0)=3$. For a nonzero integer $h$, direct integration gives

\gamma(h)=\frac{3}{2\pi}\int_{-\pi}^{\pi}e^{ih\lambda}\,d\lambda =\frac{3}{2\pi}\left[\frac{e^{ih\lambda}}{ih}\right]_{-\pi}^{\pi} =\frac{3}{2\pi}\frac{e^{ih\pi}-e^{-ih\pi}}{ih} =\frac{3\sin(h\pi)}{\pi h}=0.

Here $\sin(h\pi)=0$ because $h$ is an integer. Therefore the covariance is $3$ at lag zero and $0$ at every nonzero lag, as for white noise with variance $3$.

Check: the lag-zero integral gives variance $3$, and the nonzero-lag integral vanishes. Thus the total mass and the full covariance sequence agree with the stated density.

Worked Example: A Trigonometric Density

Let $f(\lambda)=\tfrac{1}{2\pi}(2+\cos\lambda)$ on $[-\pi,\pi)$. This is nonnegative because $2+\cos\lambda\geq1$, and it is even. Compute all the covariances.

The total mass is $\tfrac{1}{2\pi}\int_{-\pi}^{\pi}2\,d\lambda+\tfrac{1}{2\pi}\int_{-\pi}^{\pi}\cos\lambda\,d\lambda=2+0=2$, so $\gamma(0)=2$. For an integer $h$, use $\cos\lambda=\tfrac12(e^{i\lambda}+e^{-i\lambda})$:

\gamma(h)=\frac{1}{2\pi}\int_{-\pi}^{\pi}2e^{ih\lambda}\,d\lambda +\frac{1}{4\pi}\int_{-\pi}^{\pi} \left(e^{i(h+1)\lambda}+e^{i(h-1)\lambda}\right)\,d\lambda.

For any integer $k$, $\int_{-\pi}^{\pi}e^{ik\lambda}\,d\lambda=2\pi$ when $k=0$, and is $0$ when $k\ne0$. The first integral therefore contributes $2$ if $h=0$ and $0$ otherwise. The second contributes $\tfrac12$ if $h=-1$ or $h=1$, and $0$ for all other integer lags. At $h=0$, both exponents in the second integral have nonzero frequency, so that term contributes zero. Hence $\gamma(0)=2$, $\gamma(1)=\gamma(-1)=\tfrac12$, and $\gamma(h)=0$ for $|h|>1$.

Check: the sequence has total variance $2$ at lag zero, and the density is nonnegative and even. The proposition therefore confirms that the computed sequence is a valid real covariance sequence.

Worked Example: Density Supported on a Frequency Interval

Consider the even density $f(\lambda)=\tfrac1\pi$ for $-\tfrac\pi2\leq\lambda\leq\tfrac\pi2$ and $f(\lambda)=0$ elsewhere in $[-\pi,\pi)$. The endpoints of the interval are included; their inclusion does not change any integral. Find $\gamma(h)$.

The total mass is $\int_{-\pi/2}^{\pi/2}\tfrac1\pi\,d\lambda=\tfrac1\pi\cdot\pi=1$, so $\gamma(0)=1$. For a nonzero integer $h$,

\gamma(h)=\frac1\pi\int_{-\pi/2}^{\pi/2}e^{ih\lambda}\,d\lambda =\frac{1}{\pi ih}\left(e^{ih\pi/2}-e^{-ih\pi/2}\right) =\frac{2\sin(h\pi/2)}{\pi h}.

In particular, $\gamma(1)=\tfrac2\pi$, $\gamma(2)=0$, and $\gamma(3)=-\tfrac{2}{3\pi}$, since $\sin(\pi/2)=1$, $\sin(\pi)=0$, and $\sin(3\pi/2)=-1$. The general formula also gives $\gamma(-h)=\gamma(h)$: both the numerator and denominator change sign when $h$ changes sign.

Check: the direct lag-zero integral is $1$, matching the variance. For $h=2$, the integral formula gives $2\sin(\pi)/ (2\pi)=0$, confirming the stated zero at that lag.

Normalization and Interpretation

The most common error is to switch conventions halfway through a calculation. Under the convention used here, $F(d\lambda)=f(\lambda)\,d\lambda$, so the density integrates to the variance. For white noise with variance $3$, using the incorrect constant $f(\lambda)=3$ would give $\gamma(0)=\int_{-\pi}^{\pi}3\,d\lambda=6\pi$, not $3$. The correct density is $3/(2\pi)$. Some references instead use normalized measure $d\lambda/(2\pi)$; their density values differ by a factor of $2\pi$. A formula from that convention cannot be combined with the ordinary-$d\lambda$ integral without converting the density.

The integral also gives a practical validity check: any proposed density must be nonnegative almost everywhere, and its integral must equal the process variance. For a real process, an even density produces real, symmetric covariances. These checks do not require the density to be smooth; the interval-supported example has jumps and still yields the covariance integral at every integer lag.

Check Your Understanding

Use the convention $F(d\lambda)=f(\lambda)\,d\lambda$ throughout.

If $f(\lambda)=\tfrac{5}{2\pi}$ on $[-\pi,\pi)$, what are $\gamma(0)$ and $\gamma(h)$ for nonzero integer $h$?
For $f(\lambda)=\tfrac{1}{2\pi}(1+\cos(2\lambda))$, compute $\gamma(0)$, $\gamma(2)$, and $\gamma(-2)$.
Why does a nonnegative even integrable density produce a positive-definite covariance sequence?
A proposed density integrates to $4$. What must the variance be under this convention?
Does a density for only the absolutely continuous part of a spectral measure necessarily recover the entire covariance sequence? Explain.

Spectral Densities and Fourier Inversion

What You'll Learn