Bochner’s Theorem for Continuous-Time Covariances

From Positive Definiteness to a Spectral Measure

For a weakly stationary process in continuous time, covariance depends on the time difference. The resulting function cannot be arbitrary: every finite collection of time points must produce a valid covariance matrix. Bochner’s theorem says that, when the covariance function is continuous, this condition is exactly what is needed for it to be the Fourier transform of a finite nonnegative measure. The measure may have atoms, a density, or both; the theorem itself does not require a density.

Let \((X_t)_{t\ge 0}\) be a real-valued, second-order weakly stationary process on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Its covariance at lag \(t\) is denoted \(C(t)\), extended to all \(t\in\mathbb{R}\) by \(C(-t)=C(t)\). A covariance function is positive definite: for any integer \(m\ge 1\), times \(t_1,\ldots,t_m\in\mathbb{R}\), and complex numbers \(z_1,\ldots,z_m\), the sum \(\sum_{j,k=1}^m z_j\overline{z_k}C(t_j-t_k)\) is nonnegative. More generally, the theorem concerns any continuous, real-valued, even function \(C\) with this property.

We use the Fourier convention \(C(t)=\int_{\mathbb{R}}e^{it\omega}\,F(d\omega)\). Here \(F\) is a measure on the frequency line, not necessarily a density. With this convention, the inverse Fourier formula for an integrable function \(h\), when it applies, has the factor \(1/(2\pi)\) and the exponential \(e^{-it\omega}\).

Proof. First suppose such a measure \(F\) exists. For arbitrary \(m,t_1,\ldots,t_m,z_1,\ldots,z_m\), finite summation can be moved inside the integral, giving

Thus \(C\) is positive definite. Also \(C(0)=F(\mathbb{R})\), and symmetry of \(F\) makes the transform real and even.

For the converse, assume \(C\) is continuous, real, even, and positive definite. The two-point positive-definiteness condition implies \(|C(t)|\le C(0)\): the matrix with diagonal entries \(C(0)\) and off-diagonal entries \(C(t)\) is positive semidefinite, so its determinant \(C(0)^2-C(t)^2\) is nonnegative. If \(C(0)=0\), this bound gives \(C(t)=0\) for every \(t\), represented by the zero measure. Hence assume \(C(0)>0\).

Step 1: smooth the function. For \(\varepsilon>0\), put \(h_\varepsilon(t)=C(t)e^{-\varepsilon t^2/2}\). This function is integrable because \(|C(t)|\le C(0)\). It is also positive definite. Indeed, the Gaussian factor has the integral representation \(e^{-\varepsilon u^2/2}=\int e^{iux}G_\varepsilon(dx)\), where \(G_\varepsilon\) is the centered Gaussian probability measure with variance \(\varepsilon\). In a finite positive-definiteness sum for \(h_\varepsilon\), insert this representation and move the finite sum inside the integral. For each \(x\), the resulting sum is the positive-definiteness sum for \(C\) with coefficients \(z_j e^{it_jx}\); it is nonnegative. Integrating preserves nonnegativity.

Define \(g_\varepsilon(\omega)=\frac{1}{2\pi}\int_{\mathbb{R}}h_\varepsilon(t)e^{-it\omega}\,dt\). We show it is nonnegative rather than assuming that a Fourier transform of a positive function must be nonnegative. The kernel \(h_\varepsilon(u-v)e^{-i\omega(u-v)}\) is positive definite: multiplying each coefficient at \(u\) by \(e^{-i\omega u}\) reduces its quadratic form to that of \(h_\varepsilon\). Approximation by Riemann sums therefore gives, for \(T>0\),

Divide by \(2T\) and let \(T\) increase to infinity. The weight \((1-|s|/(2T))\mathbf{1}_{\{|s|\le2T\}}\) is bounded by \(1\) and tends to \(1\); since \(h_\varepsilon\) is integrable, dominated convergence gives \(\int h_\varepsilon(s)e^{-i\omega s}\,ds\ge0\). Consequently \(g_\varepsilon(\omega)\ge0\).

Step 2: show these densities have the right mass and transform. The Gaussian integral identity \(\int e^{-x^2/2}e^{iax}\,dx=\sqrt{2\pi}e^{-a^2/2}\) gives, for \(\delta>0\),

The right side tends to \(h_\varepsilon(0)=C(0)\) as \(\delta\) decreases to zero: it is the average of the continuous function \(h_\varepsilon\) against a centered Gaussian density whose variance tends to zero. On the left, the factors \(e^{-\delta\omega^2/2}\) increase to \(1\), so monotone convergence proves \(\int g_\varepsilon(\omega)\,d\omega=C(0)\). In particular, \(\mu_\varepsilon(d\omega)=g_\varepsilon(\omega)\,d\omega\) is a finite measure.

For fixed \(t\), insert the defining integral for \(g_\varepsilon\) into \(\int e^{it\omega}e^{-\delta\omega^2/2}g_\varepsilon(\omega)\,d\omega\). Fubini’s theorem applies because \(h_\varepsilon\) and the Gaussian factor are integrable. The result is the average of \(h_\varepsilon(s)\) against the Gaussian density with mean \(t\) and variance \(\delta\). As \(\delta\) decreases to zero this tends to \(h_\varepsilon(t)\), by continuity. On the original side, dominated convergence applies because \(g_\varepsilon\) is integrable. Therefore \(\int e^{it\omega}\,\mu_\varepsilon(d\omega)=h_\varepsilon(t)\).

Step 3: take a limit of the measures. The measures \(\mu_\varepsilon\) have mass \(C(0)\), and their transforms \(h_\varepsilon(t)\) tend to \(C(t)\) as \(\varepsilon\) decreases to zero. We need tightness: mass cannot escape to arbitrarily large frequencies. For \(a>0\), averaging \(1-\cos(t\omega)\) over \([-a,a]\) gives \(1-\sin(a\omega)/(a\omega)\), with the ratio interpreted as \(1\) at \(\omega=0\). If \(|\omega|>2/a\), then \(|\sin(a\omega)/(a\omega)|<1/2\), so this average exceeds \(1/2\). It follows that

For any sequence \(\varepsilon_n\to0\), the integrand converges uniformly on each bounded interval to \(C(0)-C(t)\). Continuity at zero makes the right side tend to zero as \(a\) decreases to zero, uniformly for all sufficiently large \(n\). The finitely many remaining finite measures are individually tight. Thus the sequence is tight. By the standard Prokhorov compactness theorem for finite measures of fixed mass on \(\mathbb{R}\), a subsequence converges weakly to a finite nonnegative measure \(F\). This compactness theorem is used here as a standard result; its proof is beyond this tutorial. Weak convergence preserves integrals of bounded continuous functions, so, for each \(t\),

Taking \(t=0\) gives \(F(\mathbb{R})=C(0)\).

It remains to check uniqueness. Suppose two finite nonnegative measures \(F_1,F_2\) have the same transform, and let \(\sigma=F_1-F_2\), a finite signed measure with zero transform. For every \(\delta>0\), its convolution with a centered Gaussian density of variance \(\delta\) is zero: the Gaussian Fourier integral expresses this convolution as an integral of the zero transform multiplied by \(e^{-\delta t^2/2}\). Hence, for every continuous function \(\varphi\) with compact support, \(\int(\varphi*q_\delta)(x)\,\sigma(dx)=0\), where \(q_\delta\) is that Gaussian density. As \(\delta\) decreases to zero, \(\varphi*q_\delta\) converges uniformly to \(\varphi\); finiteness of the total variation of \(\sigma\) then gives \(\int\varphi\,d\sigma=0\). Continuous compactly supported test functions determine finite measures, so \(F_1=F_2\). Finally, reflecting \(F\) about zero preserves its transform because \(C\) is even. Uniqueness therefore implies \(F\) is symmetric. This proves both directions and all the stated properties. \(\square\)

Reading and Constructing the Measure

The theorem is an existence-and-uniqueness statement, not an assertion that every spectral measure has a density. When a density does exist, the stated Fourier convention fixes its constants. The examples below illustrate three different measure forms and verify the transform directly.

What the Characterization Does—and Does Not—Say

For a covariance function, the theorem gives \(C(0)=F(\mathbb{R})\): total spectral mass equals variance. Positive definiteness is the essential restriction because it requires every finite covariance matrix to be positive semidefinite. Ordinary pointwise nonnegativity is not enough. Continuity is also a hypothesis of the theorem proved here; it cannot be silently dropped from this version.

Common pitfall: confusing a positive function with a positive-definite function. The continuous function \(C(t)=1+t^2\) is positive at every \(t\), but it cannot be a covariance function. At times \(0\) and \(1\), its proposed covariance matrix would be \(\begin{bmatrix}1&2\\2&1\end{bmatrix}\), whose determinant is \(1-4=-3\). A covariance matrix must be positive semidefinite, so this candidate fails positive definiteness. The correction is to test the full quadratic-form condition, not just the sign of \(C(t)\).

A second frequent error is changing Fourier conventions midway through a calculation. Under the convention used here, a density is recovered with a factor \(1/(2\pi)\) in the inverse transform. For example, the Gaussian density in the preceding example integrates to \(1\); dropping its factor \(1/(2\sqrt{\pi})\) would change both the total mass and the covariance at lag zero.