Positive Definiteness of Characteristic Functions

From Exponential Sums to Positive Matrices

A characteristic function is an average of complex exponentials. Although its values may be complex or negative, averaging does not erase a useful structure: when the function is evaluated at all pairwise differences among finitely many real numbers, the resulting matrix is positive semidefinite. This property comes directly from the nonnegativity of a squared absolute value, so it requires no density, moments, or special form of the distribution.

This structure provides a practical check on proposed characteristic functions. A single finite matrix that is not positive semidefinite rules out a candidate, even if the candidate has some familiar features such as being continuous and equal to one at zero.

Definition: A function $g:\mathbb{R}\to\mathbb{C}$ is positive definite if, for every positive integer $m$, every choice $t_1,\ldots,t_m\in\mathbb{R}$, and every choice $z_1,\ldots,z_m\in\mathbb{C}$, $$ \sum_{j=1}^{m}\sum_{k=1}^{m}\overline{z_j}\,g(t_j-t_k)\,z_k \geq 0. $$ Here the inequality means that the displayed quantity is real and nonnegative.

Repeated points $t_j$ are allowed, and the coefficients $z_j$ can be arbitrary complex numbers. Equivalently, for every finite list of points the matrix $M$ with entries $M_{jk}=g(t_j-t_k)$ must be positive semidefinite: its quadratic form $\overline{z}^{\,T}Mz$ is nonnegative for every complex vector $z$.

Why Every Characteristic Function Is Positive Definite

Let $X$ be a real-valued random variable with characteristic function $\varphi(t)=E[e^{itX}]$. For any finite set of points and coefficients, expand a squared absolute value. The terms in the expansion involve precisely the pairwise differences $t_j-t_k$.

Theorem: Every characteristic function $\varphi(t)=E[e^{itX}]$ is positive definite. In particular, for every $m$, every $t_1,\ldots,t_m\in\mathbb{R}$, and every $z_1,\ldots,z_m\in\mathbb{C}$, the matrix $M$ defined by $M_{jk}=\varphi(t_j-t_k)$ is positive semidefinite.

Proof. Fix $m$, the points $t_1,\ldots,t_m$, and the coefficients $z_1,\ldots,z_m$. Since $|e^{itX}|=1$, the finite sum of exponentials below is bounded in absolute value by $\sum_j|z_j|$. Its squared absolute value therefore has a finite expectation. Expand it and take expectations:

\begin{aligned} E\left[\left|\sum_{j=1}^{m}\overline{z_j}e^{it_jX}\right|^2\right] &=E\left[\left(\sum_{j=1}^{m}\overline{z_j}e^{it_jX}\right) \left(\sum_{k=1}^{m}z_ke^{-it_kX}\right)\right]\\ &=\sum_{j=1}^{m}\sum_{k=1}^{m} \overline{z_j}z_k E\left[e^{i(t_j-t_k)X}\right]\\ &=\sum_{j=1}^{m}\sum_{k=1}^{m} \overline{z_j}\varphi(t_j-t_k)z_k. \end{aligned}

The left side is the expectation of a nonnegative real random variable, so it is real and nonnegative. The right side is the quadratic form in the definition of positive definiteness. This proves the claim for every finite choice of points and coefficients. $\square$

The matrix is also Hermitian, as a positive semidefinite matrix must be. Indeed, characteristic functions satisfy $\varphi(-u)=\overline{\varphi(u)}$, since $e^{-iuX}=\overline{e^{iuX}}$. Thus $M_{kj}=\varphi(t_k-t_j)=\overline{\varphi(t_j-t_k)}=\overline{M_{jk}}$. Its diagonal entries are all $\varphi(0)=1$.

The Two-Point Test and the Modulus Bound

The simplest nontrivial matrix test uses two points, $0$ and $h$. Write $c=\varphi(h)$. With the row and column convention $M_{jk}=\varphi(t_j-t_k)$, the matrix is

M= \begin{pmatrix} 1 & \overline{c}\\ c & 1 \end{pmatrix}.

Positive semidefiniteness requires every quadratic form from this matrix to be nonnegative. In particular, choose the coefficient vector $z=(1,-c)^T$. Direct substitution gives

\begin{aligned} \overline{z}^{\,T}Mz &=1+|c|^2-\overline{c}c-\overline{c}c\\ &=1-|c|^2. \end{aligned}

Because this value must be nonnegative, $|c|\leq 1$. Since $h$ was arbitrary, this recovers the modulus bound $|\varphi(h)|\leq 1$ for every real $h$. This calculation also illustrates why the coefficient choice and matrix convention matter: for this displayed matrix, the choice $z=(1,-c)^T$ gives $1-|c|^2$, whereas replacing $-c$ with $-\overline c$ does not in general do so.

Worked Example: A Two-Point Matrix for a Bernoulli Variable

Let $X$ equal $1$ with probability $0.3$ and $0$ with probability $0.7$. Its characteristic function is $\varphi(t)=0.7+0.3e^{it}$. Use the points $0$ and $\pi$ to check the corresponding matrix.

First, $c=\varphi(\pi)=0.7+0.3e^{i\pi}=0.7-0.3=0.4$. Since $c$ is real, the matrix is

M= \begin{pmatrix} 1 & 0.4\\ 0.4 & 1 \end{pmatrix}.

The vectors $(1,1)^T$ and $(1,-1)^T$ are eigenvectors, with eigenvalues $1.4$ and $0.6$, respectively. Both eigenvalues are positive, so the matrix is positive definite. Equivalently, the two-point test gives determinant $1-(0.4)^2=0.84>0$, with positive diagonal entries. This is a check for these two points; the general theorem guarantees the condition for every finite selection of points.

Checking Larger Matrices

The positive-definiteness requirement applies to every finite number of points, not just two. Larger matrices can reveal structure that the two-point test cannot show. A matrix may be positive semidefinite without being positive definite: a zero eigenvalue is permitted because the defining quadratic form need only be nonnegative, not strictly positive.

Worked Example: A Three-Point Matrix That Is the Identity

Let $X$ be uniform on $\{0,1,2\}$. Its characteristic function is $\varphi(t)=(1+e^{it}+e^{2it})/3$. Choose $t_1=0$, $t_2=2\pi/3$, and $t_3=4\pi/3$.

For a diagonal entry, the argument is zero and $\varphi(0)=1$. For an off-diagonal entry, the argument is one of $\pm2\pi/3$ or $\pm4\pi/3$. Each is congruent modulo $2\pi$ to either $2\pi/3$ or $-2\pi/3$. At either such argument, the three terms in the numerator are the three cube roots of unity, whose sum is zero. Thus every off-diagonal entry is zero, and

M= \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}.

For any $z\in\mathbb{C}^3$, the quadratic form is $\overline z^{\,T}Mz=|z_1|^2+|z_2|^2+|z_3|^2\geq0$. The matrix condition is therefore satisfied, in this case strictly for every nonzero $z$.

Worked Example: A Singular Matrix from a Valid Characteristic Function

Let $X$ take the values $-1$ and $1$, each with probability $1/2$. Then $\varphi(t)=\cos t$. Choose $t_1=0$ and $t_2=\pi$. Since $\varphi(\pi)=\cos\pi=-1$, the matrix is

M= \begin{pmatrix} 1&-1\\ -1&1 \end{pmatrix}.

For $z=(z_1,z_2)^T$, its quadratic form is $|z_1-z_2|^2$, which is nonnegative. The eigenvalues are $0$ and $2$, so the matrix is positive semidefinite but singular. The negative off-diagonal entries do not conflict with positive semidefiniteness: it is the whole quadratic form, not each entry separately, that must be nonnegative.

Using the Condition to Reject a Candidate

Positive definiteness is a necessary structural condition for a characteristic function. A failed finite matrix test is decisive: the function cannot be a characteristic function. Passing a few selected tests, however, only shows that those particular matrices satisfy the condition; it does not establish positive definiteness for every finite selection of points.

Worked Example: A Continuous, Normalized Candidate That Fails

Consider $g(t)=1+t^2$. It is real, even, continuous, and satisfies $g(0)=1$, but those properties do not ensure it can be a characteristic function. Choose the points $0$ and $1$. The matrix is

M= \begin{pmatrix} g(0)&g(-1)\\ g(1)&g(0) \end{pmatrix} = \begin{pmatrix} 1&2\\ 2&1 \end{pmatrix}.

Taking $z=(1,-1)^T$ gives $\overline z^{\,T}Mz=1-2-2+1=-2<0$. Thus this matrix is not positive semidefinite, so $g$ cannot be a characteristic function. One counterexample matrix is enough to rule it out.

Common Pitfalls and Why the Condition Matters

A common mistake is to confuse positive definiteness with pointwise nonnegativity. A characteristic function need not have nonnegative values: in the preceding example, the valid function $\cos t$ takes the value $-1$ at $t=\pi$. Positive definiteness instead concerns a joint quadratic-form condition across selected function values.

A second pitfall is to treat the matrix entries as if they were all required to be nonnegative real numbers. They need not be; they may be complex, and even real entries may be negative. The requirements are that the matrix be Hermitian and that every complex quadratic form $\overline z^{\,T}Mz$ be real and nonnegative. In a two-point check, keeping track of which entry is $\varphi(h)$, which is $\varphi(-h)$, and where the conjugates appear is essential.

The matrix condition matters because it turns a basic identity for exponential sums into a family of algebraic restrictions on characteristic functions. The two-point case recovers the modulus bound, while larger matrices can expose incompatibilities that are not visible from individual values alone. These tests are necessary checks, not a shortcut that replaces checking all finite choices of points.

Key takeaway: For any finite set of real points, a characteristic function evaluated at their pairwise differences forms a positive semidefinite Hermitian matrix. This follows by expanding the expectation of a squared absolute value, and every finite matrix must pass the test.

Check Your Understanding

Try these questions using the matrix convention $M_{jk}=\varphi(t_j-t_k)$.

For points $0$ and $h$, write the two-by-two matrix in terms of $c=\varphi(h)$. Which coefficient vector gives the quadratic form $1-|c|^2$?
Why is the squared exponential sum used in the proof integrable even when no moment of $X$ has been assumed?
Can a characteristic function take a negative value? Give the example from this tutorial and explain why that does not violate positive definiteness.
What does a single finite matrix with a negative quadratic form imply about a candidate function?
Does checking positive semidefiniteness for one two-point matrix prove that a function is positive definite? Explain.