Why Positive Definiteness Characterizes Characteristic Functions
A characteristic function is not just a continuous function that happens to equal 1 at zero. Its values must fit together in a way that could arise from averaging the complex exponentials \(e^{itX}\) over a probability distribution. Positive definiteness expresses this compatibility through finite quadratic forms. Bochner’s theorem says that, on the real line, continuity and normalization complete the characterization.
The theorem is useful in two directions. A probability model automatically gives a positive-definite characteristic function; conversely, a function satisfying the required conditions is guaranteed to come from some probability distribution. The converse is substantial: it turns a set of analytic checks into a guarantee that a probability law exists.
\(\displaystyle \sum_{j=1}^{m}\sum_{k=1}^{m} z_j\overline{z_k}\,\phi(t_j-t_k)\geq 0.\)
The inequality means that the sum is real and nonnegative for every such choice. Equivalently, every matrix with entries \(\phi(t_j-t_k)\) is positive semidefinite.
In this course, the characteristic function of a real-valued random variable \(X\) is \(\phi_X(t)=E[e^{itX}]\). The differences \(t_j-t_k\) in the definition match the differences that appear when a squared absolute value is expanded.
Why the Conditions Are Necessary
Suppose \(\phi(t)=E[e^{itX}]\) for a real-valued random variable \(X\). Normalization follows by setting \(t=0\): \(\phi(0)=E[1]=1\). Continuity follows from the dominated convergence theorem: if \(t_r\to t\), then \(e^{it_rX}\to e^{itX}\) for every finite \(X\), and \(|e^{it_rX}|=1\), an integrable bound.
For positive definiteness, take any \(m\), \(t_1,\ldots,t_m\), and \(z_1,\ldots,z_m\). Expanding the squared absolute value gives
The equality follows by substituting \(\phi(t_j-t_k)=E[e^{i(t_j-t_k)X}]\) and bringing the finite sum inside the expectation. A squared absolute value is nonnegative, so its expectation is nonnegative. This proves that every characteristic function has all three stated properties. \(\square\)
Why the Conditions Are Sufficient
The converse is more involved: starting from a function, we must produce a probability distribution whose characteristic function is that function. The key idea is to smooth the candidate with a Gaussian factor. The smoothed function is integrable, which lets us construct a nonnegative density by an inverse Fourier integral. As the smoothing is removed, tightness ensures that some subsequence of the resulting probability distributions converges to a probability distribution.
Here are the details. Two standard facts used in the construction are recorded explicitly. First, the product of two positive-definite functions is positive definite: their finite matrices are positive semidefinite, and the entrywise product of two such matrices is positive semidefinite. Second, a tight sequence of probability distributions on \(\mathbb{R}\) has a weakly convergent subsequence. The latter follows by selecting a subsequence on which the distribution functions converge at every rational number, using a diagonal argument; tightness ensures the limiting distribution function has limits 0 and 1 at the two ends of the line.
For \(\varepsilon>0\), set \(\psi_\varepsilon(t)=\phi(t)e^{-\varepsilon t^2/2}\). The Gaussian factor is a characteristic function, hence positive definite; the product is positive definite. Also, \(\psi_\varepsilon\) is continuous, \(\psi_\varepsilon(0)=1\), and it is integrable because positive definiteness and normalization imply \(|\phi(t)|\leq 1\).
Define \(g_\varepsilon(x)=(2\pi)^{-1}\int_{\mathbb{R}}e^{-itx}\psi_\varepsilon(t)\,dt\). Positive definiteness implies \(g_\varepsilon(x)\geq0\). Indeed, its defining integral is the limit, after division by a positive constant, of the nonnegative quadratic forms obtained by integrating \(\psi_\varepsilon(s-t)h(s)\overline{h(t)}\), with \(h(s)=e^{-isx}e^{-s^2/(2R^2)}\), as \(R\to\infty\). Dominated convergence applies because \(\psi_\varepsilon\) is integrable.
For \(\delta>0\), Gaussian integration and Fubini’s theorem give \(\int e^{-\delta x^2/2}g_\varepsilon(x)\,dx=(2\pi\delta)^{-1/2}\int\psi_\varepsilon(t)e^{-t^2/(2\delta)}\,dt\). As \(\delta\downarrow0\), the right side tends to \(\psi_\varepsilon(0)=1\), by continuity and the Gaussian approximate-identity property. Since \(g_\varepsilon\geq0\), monotone convergence shows \(\int g_\varepsilon=1\). Applying the same Gaussian regularization to \(\int e^{itx}g_\varepsilon(x)\,dx\) shows it equals \(\psi_\varepsilon(t)\): the regularized integral is the convolution of \(\psi_\varepsilon\) with a centered Gaussian density of variance \(\delta\), evaluated at \(t\), and tends to \(\psi_\varepsilon(t)\).
Let \(\mu_\varepsilon\) be the probability distribution with density \(g_\varepsilon\). These distributions are tight as \(\varepsilon\downarrow0\). In fact, for any probability distribution with characteristic function \(\psi\) and any \(a>0\), \(\displaystyle P(|X|\geq2/a)\leq\frac{2}{2a}\int_{-a}^{a}(1-\operatorname{Re}\psi(t))\,dt\). The bound follows by integrating \(1-\operatorname{Re}\psi(t)\): the resulting expectation is \(E[1-\sin(aX)/(aX)]\), with the ratio defined as 1 at zero; when \(|aX|\geq2\), the bracket is at least \(1/2\). For \(\psi_\varepsilon\), the integral tends to the corresponding integral for \(\phi\), by dominated convergence. Continuity of \(\phi\) at zero makes that limiting integral, divided by \(a\), arbitrarily small for small \(a\). Thus the distributions are tight for all sufficiently small \(\varepsilon\).
Choose \(\varepsilon_r\downarrow0\). Tightness gives a subsequence of \(\mu_{\varepsilon_r}\) converging weakly to a probability distribution \(\mu\). Since \(x\mapsto e^{itx}\) is bounded and continuous, weak convergence gives convergence of the characteristic functions at every \(t\). But \(\psi_{\varepsilon_r}(t)=\phi(t)e^{-\varepsilon_r t^2/2}\to\phi(t)\), so \(\mu\) has characteristic function \(\phi\).
This proves sufficiency and completes the characterization. The proof establishes existence; it does not need to establish uniqueness of the distribution associated with \(\phi\). \(\square\)
Worked Applications
Worked Example: A Three-Point Distribution
Consider \(\phi(t)=(1+2\cos t)/3\). Check the theorem’s conditions and identify a distribution with this characteristic function.
First, \(\phi\) is continuous and \(\phi(0)=(1+2)/3=1\). To verify positive definiteness, write \[ \phi(t)=\tfrac13(1+e^{it}+e^{-it}). \] For any \(t_1,\ldots,t_m\) and \(z_1,\ldots,z_m\), the defining quadratic form becomes \[ \tfrac13\left(\left|\sum_j z_j\right|^2+ \left|\sum_j z_j e^{it_j}\right|^2+ \left|\sum_j z_j e^{-it_j}\right|^2\right)\geq0. \] Thus \(\phi\) is positive definite. It is the characteristic function of a random variable that takes values \(0,1,-1\), each with probability \(1/3\): the expectation of \(e^{itX}\) is exactly \((1+e^{it}+e^{-it})/3\). Bochner’s theorem guarantees existence from the conditions alone; the explicit distribution also verifies the result directly.
Worked Example: A Squared Sinc Function
Define \(\phi(0)=1\) and, for \(t\neq0\), let \(\phi(t)=[\sin(t/2)/(t/2)]^2\). Show that it is a characteristic function and identify its distribution.
The limit \(\sin u/u\to1\) as \(u\to0\) shows continuity at zero; continuity elsewhere is immediate. Normalization holds by definition. To verify positive definiteness, let \(U\) be uniform on \([-1/2,1/2]\). Direct integration gives \[ E[e^{itU}]=\int_{-1/2}^{1/2}e^{itu}\,du=\frac{\sin(t/2)}{t/2}, \] with value 1 at \(t=0\). If \(V\) is an independent copy of \(U\), then the characteristic function of \(U-V\) is \[ E[e^{it(U-V)}]=E[e^{itU}]E[e^{-itV}] =\left|\frac{\sin(t/2)}{t/2}\right|^2=\phi(t). \] Therefore \(\phi\) is positive definite and, by Bochner’s theorem, is a characteristic function. The difference \(U-V\) has density \(1-|x|\) for \(-1\leq x\leq1\), and density zero otherwise: at difference \(x\), the overlap of the two unit-length intervals has length \(1-|x|\) when \(|x|\leq1\).
Worked Example: A Normalized Function That Fails
Consider \(\phi(t)=1-t^2\). It is continuous and \(\phi(0)=1\), but it cannot be a characteristic function.
Positive definiteness with \(m=2\), \(t_1=0\), \(t_2=t\), and coefficients \(1\) and \(-1\) implies \[ 2\{1-\operatorname{Re}\phi(t)\}\geq0, \] so every normalized positive-definite function must satisfy \(\operatorname{Re}\phi(t)\leq1\). Here \(\phi(2)=1-4=-3\), violating that necessary inequality. Equivalently, the \(2\) by \(2\) matrix with diagonal entries 1 and off-diagonal entries \(\phi(2)=-3\) has determinant \(1-9=-8<0\), so it is not positive semidefinite. Continuity and normalization alone do not suffice.
Common Pitfall and Why the Theorem Matters
Common mistake: “The function is continuous and equals 1 at zero, so it must be a characteristic function.” Those conditions are necessary, but the example \(1-t^2\) shows they are not sufficient. The correction is to check positive definiteness as well. In practice, a proposed function may be recognized as a known characteristic function or written as a mixture or product of characteristic functions; either route can establish positive definiteness. If neither route works, the finite-matrix condition provides a way to disprove it.
Bochner’s theorem turns positive definiteness from an abstract matrix condition into a criterion for whether a probability model exists. It is especially valuable when a candidate distribution is not yet known explicitly: once continuity, normalization, and positive definiteness are established, there is a probability distribution with that characteristic function. The theorem does not by itself provide a convenient formula for the distribution, nor does the proof require one.
Check Your Understanding
For each question, use the theorem’s conditions and the finite quadratic-form definition.
- Why is the normalization condition \(\phi(0)=1\) necessary for a characteristic function?
- Write the positive-definiteness quadratic form for \(m=2\), and explain how it gives \(|\phi(t)|\leq1\) when \(\phi(0)=1\).
- For \(\phi(t)=(1+\cos t)/2\), verify continuity and normalization, then express it as a mixture of characteristic functions.
- Why does a weakly convergent subsequence of the smoothed probability distributions have a limit whose characteristic function is \(\phi\)?
- Does Bochner’s theorem require the candidate function to be real-valued? State which of its conditions rules out a general claim that it must be real.