Why Candidate Functions Need More Than a Plausible Shape
A formula can be continuous, bounded, and equal to one at zero without being the characteristic function of any probability distribution. Testing proposed functions therefore requires several distinct checks. Some are quick necessary conditions; positive definiteness is a stronger structural requirement, and continuity completes the characterization on the real line.
For a real-valued random variable \(X\), its characteristic function is \(\varphi_X(t)=E[e^{itX}]\), for \(t\in\mathbb R\). Since \(|e^{itX}|=1\), this expectation always exists. The tests below follow from that definition or from the resulting positive-definiteness property.
Four Necessary Conditions
Every characteristic function satisfies the following conditions:
- Normalization: \(\varphi(0)=1\).
- Conjugate symmetry: \(\varphi(-t)=\overline{\varphi(t)}\) for every \(t\in\mathbb R\).
- Modulus bound: \(|\varphi(t)|\leq 1\) for every \(t\in\mathbb R\).
- Positive definiteness: the finite quadratic-form condition in the definition holds.
The first three tests are often the fastest to apply. Conjugate symmetry implies that \(\varphi(t)\) is real-valued when \(\varphi\) is even: if \(\varphi(-t)=\varphi(t)\), then \(\varphi(t)=\overline{\varphi(t)}\). Passing these quick tests does not, however, guarantee positive definiteness.
Proof. At \(t=0\), \(e^{itX}=1\), so \(\varphi_X(0)=1\). Also, $$ \varphi_X(-t)=E[e^{-itX}]=\overline{E[e^{itX}]}=\overline{\varphi_X(t)}, $$ where conjugation commutes with expectation for an integrable complex random variable. The triangle inequality for expectation gives $$ |\varphi_X(t)|=|E[e^{itX}]|\leq E[|e^{itX}|]=1. $$ Finally, for arbitrary \(m,t_1,\ldots,t_m,c_1,\ldots,c_m\), $$ \sum_{j=1}^{m}\sum_{k=1}^{m}c_j\overline{c_k}\,\varphi_X(t_j-t_k) =E\left[\sum_{j=1}^{m}\sum_{k=1}^{m}c_j\overline{c_k}e^{i(t_j-t_k)X}\right] =E\left[\left|\sum_{j=1}^{m}c_j e^{it_jX}\right|^2\right]\geq 0. $$ The finite sums can be moved inside the expectation, and the last random variable is nonnegative. This proves all four claims. \(\square\)
A Complete Characterization
The four necessary conditions do not all play the same role. Positive definiteness is the key requirement that rules out many functions that pass the elementary checks. For a full validity test, continuity must also be included.
Thus, when a candidate is continuous and normalized, positive definiteness is both necessary and sufficient. Conjugate symmetry and the modulus bound are useful additional checks, but they are already consequences of being a characteristic function. They are not substitutes for positive definiteness. Bochner’s characterization also explains why a failed necessary condition is decisive: the function cannot be a characteristic function.
Worked Tests
Worked Example: A Normalization Failure in Isolation
Consider \(\varphi(t)=\frac12 e^{-t^2}\). Determine which of the stated tests it passes and whether it can be a characteristic function.
At zero, \(\varphi(0)=\frac12\), not \(1\), so normalization fails. For every real \(t\), the value is real and even, which gives \(\varphi(-t)=\overline{\varphi(t)}\). Also, \(0<\frac12 e^{-t^2}\leq\frac12\), so the modulus bound holds.
The function \(e^{-t^2}\) is the characteristic function of \(N(0,2)\). Its positive definiteness follows from the theorem above, and multiplying a positive-definite function by the positive constant \(\frac12\) preserves positive definiteness: every defining quadratic form is simply multiplied by \(\frac12\). Thus this example passes conjugate symmetry, the modulus bound, and positive definiteness, but fails normalization. It cannot be a characteristic function of a probability distribution.
Worked Example: A Conjugate-Symmetry Failure
Consider \(\varphi(t)=e^{-t^2}e^{it^2}\). Check its normalization, modulus, and conjugate symmetry.
At zero, \(\varphi(0)=1\). Its modulus is \(|\varphi(t)|=e^{-t^2}\leq1\). However, the phase \(e^{it^2}\) does not change when \(t\) is replaced by \(-t\). In particular, for \(t=\sqrt{\pi/2}\), $$ \varphi(t)=i e^{-\pi/2} \quad\text{and}\quad \varphi(-t)=i e^{-\pi/2}, \qquad \overline{\varphi(t)}=-i e^{-\pi/2}. $$ Consequently \(\varphi(-t)\neq\overline{\varphi(t)}\) at this \(t\). The function is normalized and satisfies the modulus bound, but fails conjugate symmetry; it is not a characteristic function.
Worked Example: A Modulus-Bound Failure
Let \(\varphi(t)=(1+4t^2)e^{-t^2}\). Verify its normalization and symmetry, then test its modulus.
The function is continuous and real-valued, and it is even, so it satisfies conjugate symmetry. Also, \(\varphi(0)=1\). At \(t=1\), $$ |\varphi(1)|=(1+4)e^{-1}=\frac{5}{e}>1, $$ since \(e<5\). The modulus bound fails. Therefore the function cannot be a characteristic function, regardless of whether it might pass some other test.
Worked Example: Passing Simple Checks but Failing Positive Definiteness
Consider \(\varphi(t)=\cos(t^2)\). It is continuous, satisfies \(\varphi(0)=1\), is real and even, and has \(|\varphi(t)|\leq1\). Test positive definiteness using the three points \(0,a,2a\), where \(a=\sqrt{\pi}/2\).
For a positive-definite function, the matrix with entries \(\varphi(t_j-t_k)\) must be positive semidefinite: its quadratic form is nonnegative for every vector of coefficients. Here, \(\varphi(a)=\cos(\pi/4)=\sqrt{2}/2\) and \(\varphi(2a)=\cos(\pi)=-1\). The matrix is $$ \begin{pmatrix} 1 & \sqrt{2}/2 & -1\\ \sqrt{2}/2 & 1 & \sqrt{2}/2\\ -1 & \sqrt{2}/2 & 1 \end{pmatrix}. $$ Its determinant is $$ 1-2\left(\frac12\right)+2\left(\frac12\right)(-1)-(-1)^2=-2. $$ A positive-semidefinite matrix has nonnegative determinant. This matrix does not, so the positive-definiteness condition fails. The function passes normalization, conjugate symmetry, and the modulus bound, but is not a characteristic function.
A Practical Testing Order
A systematic order helps make a rejection precise and avoids treating one successful check as proof of validity.
Evaluate the candidate at zero. If \(\varphi(0)\neq1\), it is not a characteristic function.
Compare \(\varphi(-t)\) with \(\overline{\varphi(t)}\), and look for any \(t\) with \(|\varphi(t)|>1\). Either failure rules the candidate out.
Choose finitely many points and inspect the matrix with entries \(\varphi(t_j-t_k)\). A negative quadratic form, or a negative determinant for a Hermitian matrix, disproves positive definiteness.
If the candidate is continuous, normalized, and positive definite, Bochner’s characterization establishes that it is a characteristic function.
Common Pitfall
A frequent mistake is to argue, “The function is continuous, equals one at zero, is even, and never exceeds one in absolute value, so it must be a characteristic function.” Those checks are necessary but not sufficient. The function \(\cos(t^2)\) provides a counterexample: it passes all four of those elementary checks, yet its matrix on \(0,a,2a\) has determinant \(-2\). The correction is to test positive definiteness as well, or to establish it by a valid construction. Conversely, once continuity, normalization, and positive definiteness have been verified, the characterization supplies existence; it is not necessary to guess a density or mass function first.
Check Your Understanding
For each question, identify the relevant condition and justify your answer.
- Why does \(\varphi(0)=1\) follow directly from the definition of a characteristic function?
- A candidate satisfies normalization and conjugate symmetry, but \(|\varphi(t_0)|=1.04\) for some \(t_0\). What conclusion follows?
- Why does positive definiteness require the quadratic form to be nonnegative for every choice of points and complex coefficients, rather than for just one matrix?
- What additional conditions, beyond normalization, are needed to apply Bochner’s characterization to a candidate on the real line?
- What feature of the determinant calculation rules out \(\cos(t^2)\) as a characteristic function?