White Noise Spectra and Variance

Why a Flat Spectrum Matters

A spectrum describes how the variance of a stationary process is distributed across frequencies. For uncorrelated white noise, no frequency is distinguished by the covariance structure: its spectral density is constant. The height of that constant depends on the process variance and on the frequency convention, so stating the convention is essential when interpreting or comparing spectra.

This tutorial uses angular frequency $\omega$ measured in radians per observation, over $[-\pi,\pi]$, and defines the spectral density with a $1/(2\pi)$ factor. With this convention, integrating the spectral density over the full frequency interval gives the process variance. The result follows directly from the Fourier relationship between autocovariance and spectral density.

Definition and Frequency Convention

Definition: Let $\{X_t\}$ be a covariance-stationary process with mean $\mu$, autocovariance $\gamma(h)=\operatorname{Cov}(X_t,X_{t-h})$, and autocovariances whose Fourier series is well-defined. Under the angular-frequency convention used here, its spectral density is $$ f(\omega)=\frac{1}{2\pi}\sum_{h=-\infty}^{\infty}\gamma(h)e^{-ih\omega}, \qquad -\pi\leq\omega\leq\pi. $$ The corresponding inversion formula is $$ \gamma(h)=\int_{-\pi}^{\pi}f(\omega)e^{ih\omega}\,d\omega. $$

The spectral density here describes the covariance around the mean; the mean itself is not included as a spectral component. In particular, $\gamma(0)=\operatorname{Var}(X_t)$ is the process variance. Setting $h=0$ in the inversion formula gives the variance-area relationship:

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

A white-noise process in the second-order sense has a constant mean and autocovariance equal to $\sigma^2$ at lag zero and zero at every nonzero lag. This is an uncorrelatedness condition: it does not, by itself, require independence. A process with independent, identically distributed, finite-variance observations is one example, but independence is stronger than the condition needed for the spectrum derived here.

Deriving the White-Noise Spectrum

For white noise with variance $\sigma^2$, every term in the Fourier sum vanishes except the lag-zero term. The exponential at lag zero is $e^0=1$, so the spectral density is the same at every frequency.

\begin{aligned} f(\omega) &=\frac{1}{2\pi}\sum_{h=-\infty}^{\infty}\gamma(h)e^{-ih\omega}\\ &=\frac{1}{2\pi}\gamma(0)e^0\\ &=\frac{\sigma^2}{2\pi}. \end{aligned}

The height is therefore $\sigma^2/(2\pi)$, not $\sigma^2$. Integrating that constant across an interval of length $2\pi$ recovers the variance:

\int_{-\pi}^{\pi}f(\omega)\,d\omega =\int_{-\pi}^{\pi}\frac{\sigma^2}{2\pi}\,d\omega =\frac{\sigma^2}{2\pi}(2\pi) =\sigma^2.

This calculation also shows why the frequency interval and normalization must be kept together. The constant height is defined relative to the interval $[-\pi,\pi]$ and the $1/(2\pi)$ factor in the Fourier transform. Changing conventions changes the height or integration range, but not the underlying process variance.

Worked Example: Finding the Flat Spectrum

Suppose a hypothetical stationary process has mean 12 units and is uncorrelated across time, with variance $9$ square units. Find its spectral density and verify its total spectral area.

The mean does not enter the covariance spectrum. The autocovariance is $\gamma(0)=9$ square units and $\gamma(h)=0$ for every $h\ne0$. Thus, $$ f(\omega)=\frac{1}{2\pi}\sum_{h=-\infty}^{\infty}\gamma(h)e^{-ih\omega} =\frac{9}{2\pi} \approx 1.432 $$ square units per angular-frequency unit, at every $\omega\in[-\pi,\pi]$.

The area is $$ \int_{-\pi}^{\pi}f(\omega)\,d\omega =\frac{9}{2\pi}\bigl(\pi-(-\pi)\bigr) =\frac{9}{2\pi}(2\pi) =9 $$ square units. It matches the process variance, as required by the inversion formula at lag zero.

What “Flat” Does and Does Not Mean

A flat spectrum means that the covariance structure does not favor one frequency over another. It does not mean that an observed finite sample has equal power at every frequency, nor does it mean that the process has zero variance. A particular realization will generally have an uneven-looking periodogram; the theoretical spectral density is a population property.

Likewise, “white” refers here to zero autocovariance at nonzero lags. It does not establish independence or rule out all forms of dependence. A process can have zero correlation between observations but dependence in other features, such as their squares. The spectrum is defined through second-order properties and does not capture every possible form of dependence.

Worked Example: Recovering Variance From a Flat Density

A hypothetical process has a constant spectral density of $2.5$ square meters per angular-frequency unit on $[-\pi,\pi]$. Find its variance.

By the variance-area relationship, integrate the constant density over the full interval: $$ \operatorname{Var}(X_t) =\int_{-\pi}^{\pi}2.5\,d\omega =2.5\bigl(\pi-(-\pi)\bigr) =2.5(2\pi) =5\pi \approx 15.71 $$ square meters.

This is a variance, so its units are square meters. The result also implies that the corresponding white-noise spectral density has height $\operatorname{Var}(X_t)/(2\pi)=5\pi/(2\pi)=2.5$, consistent with the value given.

Colored Noise: A Frequency-Varying Spectrum

A process can be generated by filtering white noise, which generally introduces autocovariance across time and makes the spectrum frequency-dependent. A simple example is $Y_t=\varepsilon_t+\theta\varepsilon_{t-1}$, where $\{\varepsilon_t\}$ is uncorrelated white noise with mean zero and variance $\sigma^2$. The variance of $Y_t$ is $\sigma^2(1+\theta^2)$: the two noise terms each have variance $\sigma^2$, and their covariance is zero.

For this process, the nonzero autocovariances are $\gamma_Y(0)=\sigma^2(1+\theta^2)$, $\gamma_Y(1)=\gamma_Y(-1)=\theta\sigma^2$, and zero at all other lags. Substituting these terms into the spectral-density definition gives

\begin{aligned} f_Y(\omega) &=\frac{1}{2\pi}\left[\gamma_Y(0)+\gamma_Y(1)e^{-i\omega}+\gamma_Y(-1)e^{i\omega}\right]\\ &=\frac{\sigma^2}{2\pi}\left(1+\theta^2+\theta e^{-i\omega}+\theta e^{i\omega}\right)\\ &=\frac{\sigma^2}{2\pi}\left(1+\theta^2+2\theta\cos\omega\right). \end{aligned}

The last equality uses $e^{-i\omega}+e^{i\omega}=2\cos\omega$. Unless $\theta=0$, this spectrum is generally not flat: its height varies with frequency. Its area still equals the process variance. The cosine term integrates to zero over $[-\pi,\pi]$, leaving $\sigma^2(1+\theta^2)$.

Worked Example: Checking the Area of a Colored Spectrum

Suppose a hypothetical filtered process is $Y_t=\varepsilon_t+0.5\varepsilon_{t-1}$, where the white-noise input has variance $4$ square units. Find its spectral density at zero and at $\pi$, then verify that its total area equals its variance.

Here $\theta=0.5$ and $\sigma^2=4$, so $$ f_Y(\omega)=\frac{4}{2\pi}\left(1+0.5^2+2(0.5)\cos\omega\right) =\frac{4}{2\pi}(1.25+\cos\omega). $$ At zero frequency, $\cos(0)=1$, giving $$ f_Y(0)=\frac{4}{2\pi}(2.25)=\frac{9}{2\pi}\approx1.432. $$ At frequency $\pi$, $\cos(\pi)=-1$, giving $$ f_Y(\pi)=\frac{4}{2\pi}(0.25)=\frac{1}{2\pi}\approx0.1592. $$ The spectrum is therefore higher at zero than at $\pi$; it is not flat. The process variance is $$ \operatorname{Var}(Y_t)=4(1+0.5^2)=4(1.25)=5 $$ square units. Integrating the spectrum confirms this: $$ \begin{aligned} \int_{-\pi}^{\pi}f_Y(\omega)\,d\omega &=\frac{4}{2\pi}\left[\int_{-\pi}^{\pi}1.25\,d\omega+ \int_{-\pi}^{\pi}\cos\omega\,d\omega\right]\\ &=\frac{4}{2\pi}\left[1.25(2\pi)+0\right]\\ &=5 \end{aligned} $$ square units, because $\int_{-\pi}^{\pi}\cos\omega\,d\omega=\sin(\pi)-\sin(-\pi)=0$.

Common Pitfalls and Interpretation

A common error is to call $\sigma^2$ the height of the white-noise spectral density. That omits the $1/(2\pi)$ normalization: with the convention in this tutorial, the height is $\sigma^2/(2\pi)$, and the height multiplied by the interval length $2\pi$ equals the variance. Another error is to integrate over a different frequency range without adjusting the convention. The resulting area may then be off by a factor of $2\pi$.

Spectral density also has units: process units squared per angular-frequency unit. Radians are dimensionless mathematically, but the frequency argument is still measured in radians per observation. The integral over frequency has process units squared, as a variance must. The flat spectrum gives a useful baseline; a frequency-varying spectrum indicates that the covariance structure distributes variance unevenly across frequencies, not that the process variance itself is changing over time.

Key takeaway: Under $f(\omega)=(1/(2\pi))\sum_h\gamma(h)e^{-ih\omega}$ on $[-\pi,\pi]$, uncorrelated white noise with variance $\sigma^2$ has the constant density $\sigma^2/(2\pi)$, and $\int_{-\pi}^{\pi}f(\omega)\,d\omega=\sigma^2$. Colored processes can have nonconstant spectra while retaining this same variance-area identity.

Check Your Understanding

Use the angular-frequency convention stated in this tutorial.

A white-noise process has variance $6$. What is its spectral density at every frequency, and what is the area under it over $[-\pi,\pi]$?
Why does the mean of a covariance-stationary process not appear in the spectral-density formula used here?
A constant spectral density is $3/(2\pi)$ on $[-\pi,\pi]$. What is the process variance?
For $Y_t=\varepsilon_t+\theta\varepsilon_{t-1}$, with white-noise input variance $\sigma^2$, state the spectral density and explain why it is generally not flat.
Does a flat spectrum prove that observations are independent? Explain what property the flat spectrum establishes in this setting.